SEMI-STANDARD YOUNG TABLEAUX AND sln CRYSTALS
PETER TINGLEY (TEXED BY DAVID JORDAN)
Today we will explicitly describe the combinatorics of sln crystals in terms of semi-standard
Young tableaux. As an application, we will discuss Littlewood-Richardson (LR) rules. This roughly
follows [K2, Section 5], where citations to original sources can be found.
1. The semi-setandard Young tableau realization
Recall that a semi-standard Young tableaux of shape λ for sln is a filling of λ with the numbers
{1, . . . , n}, which is weakly increasing in rows. and strictly increasing in columns. For example,
5
3 4 5
2 2 3 3 4 4 5
1 1 1 2 2 3 3 4 .
Call the set of such tableau SSY Tn (λ). If λ has more then n rows, there are no such tableaux. In
fact, we often assume λ has at most n−1 rows, as if it has n rows then the columns of height n must
all be filled with exactly 1, 2, . . . n, and can essentially be ignored. The difference between allowing
columns of height n and not is essentially the difference between working with gln representations
and working with sln representations.
Theorem 1.1. The set SSY Tn (λ) parameterizes the sln crystal B(λ), where the highest weight λ
is expressed in terms of fundamental weights by
X
λ 7→
(λj − λj−1 )ωj .
In order for this theorem to be meaningful (beyond saying the the number of semi-standard
Young tableau of shape λ is the dimension of V (λ)), we must describe the action of the crystal
operators. Each fk will change a “k” to a “k + 1”, or else set the tableau to 0. Let us describe
the algorithm by example, rather than by formulas. Let us apply f2 to the previous example. We
begin by creating a parentheses sequence, by putting a ”(” above each ”2”, and a ”)” above each
”3”, so that they cancel as parentheses do. If ”2”s and ”3”s appear in the same column, the ”(”s
appear before the ”)”, but such brackets always cancel, so many people do not include them. Here
they will be colored red.
() ) ( () ) ( (
5
3 4 5
2 2 3 3 4 4 5
1 1 1 2 2 3 3 4
We then look for the rightmost unmatched ”)”, and change the ”2” inside to a ”3”. In the previous
example we end up with the tableau:
Date: March 18, 2011.
1
2
PETER TINGLEY (TEXED BY DAVID JORDAN)
5
3 4 5
2 3 3 3 4 4 5
1 1 1 2 2 3 3 4.
If there were no uncanceled ), f2 would kill the tableaux (i.e. send it to 0). The rules for ei are
similar: you form the same string of brackets, and change the i + 1 corresponding to the leftmost
uncanceled ( to an i.
To see why the above gives the crystal B(λ), we will show that it can be embedded in a tensor
product of many copies of the crystal B(ω1 ) corresponding to the vector representation. It is easy
to check that the crystal B(ω1 ) looks like
Fn−1
F1 2 F2
1 −
→
−→ · · · · · · −−−→ n .
Consider the map SSY Tn (λ) → B(ω1 )⊗|λ| which simply reads the entries of λ moving up columns
then right to left. For example,
(1.2)
5
3 4 5
2 2 3 3 4 4 5
1 1 1 2 2 3 3 4
↓
4 ⊗ 3 ⊗ 5 ⊗ 3 ⊗ 4 ⊗ 2 ⊗ 4 ⊗ 2 ⊗ 3 ⊗ 1 ⊗ 3 ⊗ 5 ⊗ 1 ⊗ 2 ⊗ 4 ⊗ 1 ⊗ 2 ⊗ 3 ⊗ 5.
Recall the tensor product rule for crystals from Lecture 2. This can be rephrased algebraically as
follows: Let b ∈ B(λ), c ∈ B(µ) then
(
fi (b) ⊗ c if ϕi (b) > εi (c)
(1.3)
fi (b ⊗ c) =
b ⊗ fi (c) if ϕi (b) ≤ εi (c).
Even more useful to us, it can be displayed as follows: above b, put a string of brackets )) · · · ))((· · · ((,
where the number of ) is ε(b) and the number of ( is ϕ(b). Do the same for c, and concatenate to
get one string of brackets
)) · · · (( )) · · · ((
b ⊗ c.
If the leftmost uncanceled ( is in the string above b, then fi (b ⊗ c) = fi (b) ⊗ c. Otherwise,
fi (b ⊗ c) = b ⊗ fi (c).
At this point, it should be clear that the map from (1.2) intertwines the fi defined on SSY T (λ)
with the fi defined on the tensor product of the B(ω1 ) (caution: because of poor choices of conventions, one string of brackets is actually the reverse of the other). Thus our rules define an element
of the unique closed family of crystals, and by checking the highest weight we see that it is B(λ).
In fact, the above version of the tensor product rule is probably the main reason that canceling
brackets occur so often in definitions of crystal operators.
2. Jeu de Taquin
One might want to go the other way, and produce a tableaux from a sequence. This can be done
by first creating a skew tableau from the sequence, and then performing the “Jeu de Taquin.” For
SEMI-STANDARD YOUNG TABLEAUX AND sln CRYSTALS
3
example,
2
1
3 ⊗ 1 ⊗ 2
3
→
Jeux de Taquin
−−−−−−−−−→
2
1 3 .
The “Jeu de Taquin” changes a skew semi-standard tableau to an ordinary semi-standard tableau
by the following algorithm
(1) choose some box on the boundary of the inner shape.
(2) Interchange that box with either the box above it or the box to the right. Exactly one
of these will be possible without creating something with is no longer semi-standard (i.e.
weakly increasing along rows and strictly increasing along columns).
(3) Again, interchange the same box with either the box above it or the box to the right,
whichever is possible. Continue doing this until the box is an outer corner. Then delete it.
(4) Choose another box on the boundary of the inner shape, and repeat the above procedure.
(5) Continue until you have an ordinary tableau.
It turns out the the resulting tableau does not depend on the choices made. One way to see this is
to show that performing Jeu-de-taquin commutes with crystal operators (defined on skew tableau
by the obvious generalization).
Note however that many different elements of B(ω1 )⊗N can be sent to the same tableau by this
procedure. This is because there are usually many copies of a given B(λ) in B(ω1 )⊗|λ| . A related
issue is that the rule for applying fi to a tableaux can be modified by reading the boxes in various
other orders, without changing the final operator. For instance, you can read right to left along
rows, then bottom to top.
3. Littlewood-Richardson Rules
A classical problem is to compute the Littlewood-Richardson (LR) coefficient cγλ,µ , which is the
multiplicity of V (γ) in V (λ) ⊗ V (µ). We will discuss how crystal theory and the combinatorics of
Young tableau can be used to give an answer to this question. Consider the example
γ=
,
λ=
,
µ=
,
say for sl4 . To make the rule precise, it is better to consider representations of gl4 , so that highest
weights can have columns of height 4. cγλµ will always be 0 unless |γ| = |λ| + |µ|. Representations
whose highest weights differ by adding columns of height 4 are isomorphic an sl4 representations,
but not as gln representations.
We need only count the dimension of the space of highest weight vectors of weight γ that appear.
Using crystal theory, we should count the number of highest weight elements of weight γ that appear
in B(λ) ⊗ B(µ). Such a highest weight vector must by of the form bλ ⊗ c for some c in B(µ), where
3
2
bλ = 1 1
is the highest weight element in B(λ) (highest weight elements always have this form). Since the
highest weight vector bγ is given by the tableau
3 3
2 2 2
bγ = 1 1 1 1 ,
4
PETER TINGLEY (TEXED BY DAVID JORDAN)
we must count the number of tableaux of type
filled with 1, 1, 2, 2, 3 (so that the weight
of the tensor product is right), such that the product is highest weight. There are two tableaux of
this shape:
(3.1)
2 3
1 1 2 ,
and
2 2
1 1 3 .
The first satisfies the property bλ ⊗ c is highest weight, while the second does not. So in this case
cγλ,µ = 1.
More often, LR rules are described as follow: cγλ,µ is the number of skew tableaux of shape γ/λ,
satisfying the additional condition that, if you read the entries right to left along rows, starting
with the bottom row then moving up, any initial sequence has at least as many i as i + 1, for all
i. Such tableaux are called Littlewood-Richardson tableaux. The one corresponding to the above
LR-coefficient of 1 is
2
1 2
1 1 .
The correspondence between these two versions of the rule is that row r of the LR-tableaux records
the heights of all the r appearing in the corresponding tableaux from (3.1). The condition on initial
sequences exactly translates to the tableaux from (3.1) actually being semi-standard.
References
[K2] Masaki Kashiwara. On crystal bases, Representations of groups (Banff, AB, 1994), 155-197, CMS Conf. Proc.
16, Amer. Math. Soc., Providence, RI, 1995.